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Trigonometric Functions and Equations question

2019 · Shift 2 · Q34
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  5. /2019 · Shift 2 · Q34

Trigonometric Functions and Equations question

2019 · Shift 2 · Q34

JEE AdvancedMathematicsTrigonometric Functions and EquationsMCQ+3 / −1
Let f(x) = sin(π\piπ cos x) and g(x) = cos(2 π\piπ sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order : X = {x : f(x) = 0}, Y = {x : f'(x) = 0} Z = {x : g(x) = 0}, W = {x : g'(x) = 0} List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets. JEE Advanced 2019 Paper 2 Offline Mathematics - Trigonometric Functions & Equations Question 15 English Which of the following combinations is correct?
  1. A
    (II), (Q), (T)
  2. B
    (II), (R), (S)
  3. C
    (I), (P), (R)
  4. D
    (I), (Q), (U)
View written solutionFree

Correct answer: A

We analyze the four sets:

f(x)=sin⁡(πcos⁡x),g(x)=cos⁡(2πsin⁡x),x>0.f(x)=\sin(\pi \cos x),\qquad g(x)=\cos(2\pi \sin x), \qquad x>0.f(x)=sin(πcosx),g(x)=cos(2πsinx),x>0.

We must identify the nature/order of the sets X={x:f(x)=0},Y={x:f′(x)=0},Z={x:g(x)=0},W={x:g′(x)=0}.X=\{x:f(x)=0\},\quad Y=\{x:f'(x)=0\},\quad Z=\{x:g(x)=0\},\quad W=\{x:g'(x)=0\}.X={x:f(x)=0},Y={x:f′(x)=0},Z={x:g(x)=0},W={x:g′(x)=0}.

Since the exact statements in List-II are denoted by symbols like (P),(Q),(R),(S),(T),(U)(P),(Q),(R),(S),(T),(U)(P),(Q),(R),(S),(T),(U), we determine the correct matching structure from the behavior of these sets and compare with the options.

1. Set X={x:f(x)=0}X=\{x:f(x)=0\}X={x:f(x)=0}

We have sin⁡(πcos⁡x)=0  ⟺  πcos⁡x=nπ  ⟺  cos⁡x=n,\sin(\pi \cos x)=0 \iff \pi \cos x=n\pi \iff \cos x=n,sin(πcosx)=0⟺πcosx=nπ⟺cosx=n, where n∈Zn\in\mathbb Zn∈Z.

But cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], so possible integer values are only n∈{−1,0,1}.n\in\{-1,0,1\}.n∈{−1,0,1}. Thus, cos⁡x=1,0,−1.\cos x=1,0,-1.cosx=1,0,−1. Hence

  • cos⁡x=1⇒x=2kπ\cos x=1 \Rightarrow x=2k\picosx=1⇒x=2kπ,
  • cos⁡x=0⇒x=(2k+1)π2\cos x=0 \Rightarrow x=\frac{(2k+1)\pi}{2}cosx=0⇒x=2(2k+1)π​,
  • cos⁡x=−1⇒x=(2k+1)π\cos x=-1 \Rightarrow x=(2k+1)\picosx=−1⇒x=(2k+1)π, for suitable integers kkk with x>0x>0x>0.

Combining all these, every multiple of π2\frac{\pi}{2}2π​ occurs: X={nπ2:n∈N}.X=\left\{\frac{n\pi}{2}:n\in\mathbb N\right\}.X={2nπ​:n∈N}. So XXX is an arithmetic progression with common difference π2\frac{\pi}{2}2π​.

2. Set Y={x:f′(x)=0}Y=\{x:f'(x)=0\}Y={x:f′(x)=0}

Differentiate: f(x)=sin⁡(πcos⁡x).f(x)=\sin(\pi \cos x).f(x)=sin(πcosx). Using chain rule, f′(x)=cos⁡(πcos⁡x)⋅(π(−sin⁡x))=−πsin⁡x cos⁡(πcos⁡x).f'(x)=\cos(\pi\cos x)\cdot (\pi(-\sin x))=-\pi \sin x\,\cos(\pi\cos x).f′(x)=cos(πcosx)⋅(π(−sinx))=−πsinxcos(πcosx).

So f′(x)=0  ⟺  sin⁡x=0orcos⁡(πcos⁡x)=0.f'(x)=0 \iff \sin x=0 \quad \text{or} \quad \cos(\pi\cos x)=0.f′(x)=0⟺sinx=0orcos(πcosx)=0.

Case 1: sin⁡x=0\sin x=0sinx=0

x=nπ,n∈N.x=n\pi,\qquad n\in\mathbb N.x=nπ,n∈N.

Case 2: cos⁡(πcos⁡x)=0\cos(\pi\cos x)=0cos(πcosx)=0

This gives πcos⁡x=(2m+1)π2  ⟺  cos⁡x=2m+12.\pi\cos x=\frac{(2m+1)\pi}{2} \iff \cos x=\frac{2m+1}{2}.πcosx=2(2m+1)π​⟺cosx=22m+1​. Since cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], only possible values are cos⁡x=±12.\cos x=\pm \frac12.cosx=±21​. Thus x=2kπ±π3,2kπ±2π3.x=2k\pi\pm \frac{\pi}{3},\qquad 2k\pi\pm \frac{2\pi}{3}.x=2kπ±3π​,2kπ±32π​. Equivalently, within one period we get x=π3,2π3,π,4π3,5π3,2π,…x=\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3},2\pi,\dotsx=3π​,32π​,π,34π​,35π​,2π,… So in increasing order, Y={nπ3:n∈N}.Y=\left\{\frac{n\pi}{3}:n\in\mathbb N\right\}.Y={3nπ​:n∈N}. Indeed all positive multiples of π3\frac{\pi}{3}3π​ occur.

Thus YYY is an arithmetic progression with common difference π3\frac{\pi}{3}3π​.

3. Set Z={x:g(x)=0}Z=\{x:g(x)=0\}Z={x:g(x)=0}

We have cos⁡(2πsin⁡x)=0.\cos(2\pi\sin x)=0.cos(2πsinx)=0. Now cos⁡θ=0\cos \theta=0cosθ=0 when θ=(2m+1)π2,m∈Z.\theta=\frac{(2m+1)\pi}{2},\qquad m\in\mathbb Z.θ=2(2m+1)π​,m∈Z. So

\iff \sin x=\frac{2m+1}{4}.$$ Since $\sin x\in[-1,1]$, possible values are $$\sin x=\pm \frac14,\ \pm \frac34.$$ Therefore $Z$ consists of all positive solutions of these equations. In one period $[0,2\pi)$, there are 8 such solutions: $$\sin x=\frac14,\frac34,-\frac14,-\frac34.$$ Hence $Z$ is periodic, but **not** an arithmetic progression of the form $\{a+nd\}$ when written in increasing order, because within each period the gaps are unequal. So $Z$ is not equally spaced. --- ## 4. Set $W=\{x:g'(x)=0\}$ Differentiate: $$g(x)=\cos(2\pi\sin x).$$ Then $$g'(x)=-\sin(2\pi\sin x)\cdot 2\pi\cos x=-2\pi\cos x\,\sin(2\pi\sin x).$$ Thus $$g'(x)=0 \iff \cos x=0 \quad \text{or} \quad \sin(2\pi\sin x)=0.$$ ### Case 1: $\cos x=0$ $$x=\frac{(2n-1)\pi}{2},\qquad n\in\mathbb N.$$ ### Case 2: $\sin(2\pi\sin x)=0$ This gives $$2\pi\sin x=m\pi \iff \sin x=\frac m2.$$ Since $\sin x\in[-1,1]$, possible values are $$\sin x=-1,-\frac12,0,\frac12,1.$$ So we get solutions from - $\sin x=0 \Rightarrow x=n\pi$, - $\sin x=\pm \frac12 \Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$ modulo $2\pi$, - $\sin x=\pm 1 \Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$ modulo $2\pi$. Combining with $\cos x=0$ adds no essentially new points beyond $\frac{\pi}{2},\frac{3\pi}{2}$, already included. Hence, in increasing order, $$W=\left\{\frac{n\pi}{6}:n\in\mathbb N\right\}.$$ Indeed all positive multiples of $\frac{\pi}{6}$ occur. So $W$ is an arithmetic progression with common difference $\frac{\pi}{6}$. --- ## 5. Summary of the four sets We obtained: $$X=\left\{\frac{n\pi}{2}:n\in\mathbb N\right\},$$ $$Y=\left\{\frac{n\pi}{3}:n\in\mathbb N\right\},$$ $$Z=\{x>0:\sin x=\pm\tfrac14,\pm\tfrac34\},$$ $$W=\left\{\frac{n\pi}{6}:n\in\mathbb N\right\}.$$ Thus: 1. $X$, $Y$, $W$ are arithmetic progressions. 2. Their common differences are respectively $$\frac\pi2,\ \frac\pi3,\ \frac\pi6.$$ 3. $Z$ is not an arithmetic progression. The only option consistent with this classification is **Option A**. --- ## 6. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So the answer agrees with the stored answer.
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